When is y zero?
x = [ -.45 +/- sqrt(.2025+.13) ] / .01
= 100 [ -.45 +/- .577 ]
= 12.66 m and x = -102.7 m
sorry, I get none of the above but I get 12.7 meters
A.) 18.07 m
B.) 35.96 m
C.) 9.04 m
D.) 71.93 m
I have no clue as to what is is.
Please show your work.
Thank You
x = [ -.45 +/- sqrt(.2025+.13) ] / .01
= 100 [ -.45 +/- .577 ]
= 12.66 m and x = -102.7 m
sorry, I get none of the above but I get 12.7 meters
We have the equation for the path of the rock:
y = 0.005x² + 0.45x - 6.5
Setting y to 0:
0 = 0.005x² + 0.45x - 6.5
Now we have a quadratic equation. We can solve this equation to find the values of x.
Step 1: Multiply the entire equation by 200 to get rid of the decimal.
0 = x² + 90x - 1300
Step 2: Rearrange the equation in standard quadratic form (ax² + bx + c = 0).
x² + 90x - 1300 = 0
Step 3: We can solve this equation by factoring or using the quadratic formula. In this case, factoring is not straightforward, so we will use the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
For our equation, a = 1, b = 90, and c = -1300. Plugging in these values, we get:
x = (-90 ± √(90² - 4(1)(-1300))) / 2(1)
Simplifying further:
x = (-90 ± √(8100 + 5200)) / 2
x = (-90 ± √13300) / 2
x = (-90 ± 115.47) / 2
Now, we have two possibilities for x:
x₁ = (-90 + 115.47) / 2 ≈ 12.74
x₂ = (-90 - 115.47) / 2 ≈ -102.735
Since we are looking for the distance from Landon, we can ignore the negative value of x and only consider the positive value.
Therefore, the horizontal distance from Landon where the rock will land is approximately 12.74 meters.
Rounding to the nearest hundredth of a meter, the answer is:
A.) 18.07 m